-2
$\begingroup$

$$\lim_{x\to{a}}\frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a}$$

It should be $3a$, but I can't find the way to solve it without L'Hopital.

$\endgroup$
1
  • $\begingroup$ I'd suggest rationalizing the denominator: multiply by $\sqrt{ax} + a$ on top and bottom $\endgroup$ – Stephen Donovan Apr 13 at 12:46
0
$\begingroup$

The expression can be easily expanded first: \begin{align} &\frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a} \\[7pt] =~ &\frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a} \cdot \frac{\sqrt{ax}+a}{\sqrt{ax}+a}\\[7pt] =~& \frac{(x^2-\sqrt{a^3x})(\sqrt{ax}+a)}{(\sqrt{ax}-a)(\sqrt{ax}+a)} \\[7pt] =~& \frac{x^2(\sqrt{ax}+a)-\sqrt{a^3x}(\sqrt{ax}+a)}{ax-a^2} \\[7pt] =~& \frac{x^2\sqrt{ax}+ax^2-\sqrt{a^3x}\sqrt{ax}-a\sqrt{a^3x}}{a(x-a)} \end{align}

and then reduced: \begin{align} &= \frac{x^2\sqrt{ax}+ax^2-{a^2x}-a^2\sqrt{ax}}{a(x-a)} \\[7pt] &= \frac{(x^2-a^2)\sqrt{ax}+ax(x-a)}{a(x-a)} \\[7pt] &= \frac{(x+a)\sqrt{ax}+ax}{a} \\ \end{align}

Then you can simply substitute $x=a$ to get:

\begin{align} &\lim\limits_{x\to{a}}\frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a} \\[7pt] =~& \left. \frac{(x+a)\sqrt{ax}+ax}{a} \right\vert_{x=a} \\[7pt] =~& \frac{(a+a)\sqrt{a^2}+a^2}{a} \\[7pt] =~& 2a + a \\[7pt] =~& 3a \end{align}

$\endgroup$
1
  • $\begingroup$ @sera Thank you for fixing my formatting. $\endgroup$ – CiaPan Apr 13 at 14:44
2
$\begingroup$

$$ \frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a} = \sqrt{\frac xa} (x+\sqrt{xa} +a) \to 3a $$

$\endgroup$
1
$\begingroup$

I assume $a> 0$, which seems to be implicit in your question as otherwise there is a division by $0$. Write $t=\frac{x-a}{a}$, so that you want to limit as $t\to 0$ of $$\begin{align} \frac{a^2(1+t)^2- \sqrt{a^4(1+t)}}{\sqrt{a^2(1+t)}-a} &= a\cdot \frac{(1+t)^2- \sqrt{1+t}}{\sqrt{1+t}-1}\\ &= a\cdot\sqrt{1+t}\cdot \frac{(1+t)^{3/2}- 1}{\sqrt{1+t}-1}\\ &= a\cdot\sqrt{1+t}\cdot\frac{\sqrt{1+t}+1}{(1+t)^{3/2}+1} \cdot \frac{(1+t)^{3}- 1}{(1+t)-1}\\ &= a\cdot\sqrt{1+t}\cdot\frac{\sqrt{1+t}+1}{(1+t)^{3/2}+1} \cdot (3+3t+t^2)\\ &\xrightarrow[t\to ~0]{} a\cdot 1 \cdot \frac{2}{2}\cdot 3 = \boxed{3a} \\ ~ \end{align}$$

$\endgroup$
0
$\begingroup$

Hint: if you let $y=\sqrt x$ and $b=\sqrt a$, then the problem is $$\lim_{y\rightarrow b}\frac{y^4-b^3y}{by-b^2}$$ That can be factored and some cancellation can happen so this doesn't evaluate to $\frac00$.

$\endgroup$
0
$\begingroup$

With $x=at^2,$ $$\lim\limits_{x\to{a}}\frac{x^2-\sqrt{a^3x}}{\sqrt{ax}-a} =\lim\limits_{t^2\to1}\frac{a^2t^4-\sqrt{a^4t^2}}{\sqrt{a^2t^2}-a}=a\lim\limits_{t\to1}\frac{t(t^3-1)}{\text{sgn}(a)t-1}.$$

If $a\le0$, the limit does not exist. For $a>0$,

$$a\lim_{t\to1}\frac{t^4-t}{t-1}=a\lim_{t\to1}t(t^2+t+1).$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.